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Chapter 23: Problem 6

\(\bullet\) Most people perceive light having a wavelength between 630 \(\mathrm{nm}\) and 700 \(\mathrm{nm}\) as red and light with a wavelength between 400 \(\mathrm{nm}\) and 440 \(\mathrm{nm}\) as violet. Calculate the approximate frequency ranges for (a) violet light and (b) red light.

Short Answer

Expert verified
Violet light: 6.82 to 7.5 THz; Red light: 4.29 to 4.76 THz.

Step by step solution

01

Understand the Relationship

To find the frequency of light, we use the equation relating the speed of light \( c \), wavelength \( \lambda \), and frequency \( f \): \( c = \lambda f \). The speed of light \( c \) is approximately \( 3 \times 10^8 \ \text{m/s} \).
02

Convert Wavelength from nm to m

Since the speed equation requires meters, convert the given wavelengths from nanometers to meters: \( 1 \ \text{nm} = 1 \times 10^{-9} \ \text{m} \). For violet: 400 nm becomes \( 400 \times 10^{-9} \ \text{m} \) and 440 nm becomes \( 440 \times 10^{-9} \ \text{m} \). For red: 630 nm becomes \( 630 \times 10^{-9} \ \text{m} \) and 700 nm becomes \( 700 \times 10^{-9} \ \text{m} \).
03

Calculate Frequency for Violet Light

Using the speed of light equation \( c = \lambda f \), solve for frequency \( f = \frac{c}{\lambda} \). For 400 nm: \( f = \frac{3 \times 10^8 \ \text{m/s}}{400 \times 10^{-9} \ \text{m}} = 7.5 \times 10^{14} \ \text{Hz} \). For 440 nm: \( f = \frac{3 \times 10^8 \ \text{m/s}}{440 \times 10^{-9} \ \text{m}} = 6.82 \times 10^{14} \ \text{Hz} \). The frequency range for violet light is approximately 6.82 to 7.5 THz.
04

Calculate Frequency for Red Light

Using \( f = \frac{c}{\lambda} \), calculate frequency for 630 nm: \( f = \frac{3 \times 10^8 \ \text{m/s}}{630 \times 10^{-9} \ \text{m}} = 4.76 \times 10^{14} \ \text{Hz} \). For 700 nm: \( f = \frac{3 \times 10^8 \ \text{m/s}}{700 \times 10^{-9} \ \text{m}} = 4.29 \times 10^{14} \ \text{Hz} \). The frequency range for red light is approximately 4.29 to 4.76 THz.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Wavelength
Wavelength is a fundamental concept when discussing the electromagnetic spectrum and light. It describes the distance between successive crests (the highest points) of a wave. In the context of light, the wavelength determines what color the light appears to our eyes. For example, red light has longer wavelengths compared to violet light.
  • Wavelength is usually measured in meters ( ext{m}), but for light, we commonly use nanometers ( ext{nm}), where 1 ext{nm} equals 1 x 10^{-9} ext{m}.
  • The visible light spectrum ranges approximately from 400 nm (violet) to 700 nm (red).
Understanding wavelength helps us decode the properties of light and predict its behavior, such as how it interacts with different materials and how it is perceived by the human eye.
Frequency
Frequency refers to the number of wave cycles that pass a point in one second, expressed in hertz (Hz). In the context of light, frequency helps us determine the energy of light waves.
  • Higher frequency corresponds to greater energy. Thus, violet light, which has a higher frequency, is more energetic than red light.
  • The formula to find frequency is \( f = \frac{c}{\lambda} \), where \( f \) is frequency, \( c \) is the speed of light, and \( \lambda \) is wavelength.
As frequency increases from red to violet, the energy of the light waves also increases, affecting their color and visibility.
Speed of Light
The speed of light is a constant that plays a crucial role in the study of physics and the understanding of the universe. It is \(3 \times 10^8 \text{ m/s}\) in a vacuum, and this constant speed is a key part of equations involving electromagnetic waves.
  • The relationship between the speed of light, frequency, and wavelength is given by \( c = \lambda f \).
  • This equation is essential for converting between wavelength and frequency, as done in the exercise to calculate the frequencies for red and violet light.
The speed of light is not only a fundamental constant in physics but also a bridge between wavelength and frequency, allowing us to convert and understand these two properties in relation to each other.
Visible Light
Visible light is a small portion of the electromagnetic spectrum that is detectable by the human eye. This range encompasses a variety of colors from violet to red, each with its unique wavelength and frequency.
  • Violet light has the shortest wavelength, around 400 nm, and highest frequency, while red light has the longest wavelength, around 700 nm, and lowest frequency.
  • Light within this range can be split into various colors using prisms or diffraction grating, creating a spectrum like a rainbow.
Understanding visible light helps us understand the broader electromagnetic spectrum and how energy travels in different forms throughout the universe. It also highlights the fascinating ways in which physics becomes a part of our everyday experiences, from the colors we perceive to the technology we use.

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Most popular questions from this chapter

\(\bullet\) Unpolarized light with intensity \(I_{0}\) is incident on an ideal polarizing filter. The emerging light strikes a second ideal polarizing filter whose axis is at \(41.0^{\circ}\) to that of the first. Determine (a) the intensity of the beam after it has passed through the second polarizer and (b) its state of polarization.

\(\bullet$$\bullet\) Heart sonogram. Physicians use high-frequency \((f=\) 1 MHz to 5 MHz) sound waves, called ultrasound, to image internal organs. The speed of these ultrasound waves is 1480 \(\mathrm{m} / \mathrm{s}\) in muscle and 344 \(\mathrm{m} / \mathrm{s}\) in air. We define the index of refraction of a material for sound waves to be the ratio of the speed of sound in air to the speed of sound in the material. Snell's law then applies to the refraction of sound waves. (a) At what angle from the normal does an ultrasound beam enter the heart if it leaves the lungs at an angle of \(9.73^{\circ}\) from the normal to the heart wall? (Assume that the speed of sound in the lungs is 344 \(\mathrm{m} / \mathrm{s} .\) ) (b) What is the critical angle for sound waves in air incident on muscle?

\(\bullet$$\bullet\) Two sources of sinusoidal electromagnetic waves have average powers of 75 \(\mathrm{W}\) and 150 \(\mathrm{W}\) and emit uniformly in all directions. At the same distance from each source, what is the ratio of the maximum electric field for the 150 \(\mathrm{W}\) source to that of the 75 \(\mathrm{W}\) source?

\(\bullet\) The electric field of a sinusoidal electromagnetic wave obeys the equation \(E=-(375 \mathrm{V} / \mathrm{m}) \sin [(5.97 \times\) \(10^{15} \operatorname{rad} / \mathrm{s} ) t+\left(1.99 \times 10^{7} \mathrm{rad} / \mathrm{m}\right) x ] .\) (a) What are the amplitudes of the electric and magnetic fields of this wave? (b) What are the frequency, wavelength, and period of the wave? Is this light visible to humans? (c) What is the speed of the wave?

\(\bullet\) Consider electromagnetic waves propagating in air. (a) Determine the frequency of a wave with a wavelength of (i) \(5.0 \mathrm{km},\) (ii) \(5.0 \mu \mathrm{m},\) (iii) 5.0 \(\mathrm{nm}\) . (b) What is the wavelength (in meters and nanometers) of (i) gamma rays of frequency \(6.50 \times 10^{21} \mathrm{Hz}\) , (ii) an AM station radio wave of frequency 590 \(\mathrm{kHz} ?\)
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